Recovering the Original Simplicity: Succinct and Exact Quantum Algorithm for the Welded Tree Problem

Abstract

This work revisits quantum algorithms for the well-known welded tree problem, proposing a succinct quantum algorithm based on the simple coined quantum walks. It iterates the naturally defined coined quantum walk operator for a classically precomputed number of iterations, and measures. The number of iterations is linear in the depth of the tree. The success probability of this procedure is inversely linear in the depth of the tree. Moreover, it is the same for all instances of the problem of a fixed size, therefore, we can use the exact quantum amplitude amplification subroutine to answer with probability 1. This gives an exponential speedup over any classical algorithm for the same problem. The significance of the results may be seen as follows. (i) Our algorithm is rather simple compared with the one in (Jeffery and Zur, STOC’2023), which not only breaks the stereotype that coined quantum walks can only achieve quadratic speedups over classical algorithms, but also demonstrates the power of the simplest quantum walk model. (ii) Our algorithm achieves certainty of success for the first time. Thus, it becomes one of the few examples that exhibit exponential separation between exact quantum and randomized query complexities.

Appendices

Implementing $U_\varphi$ in Lemma 2.1

In this appendix, we will implement the unitary operator $U_\varphi : {|{u,\perp }\rangle } \mapsto {|{u, \varphi (u)}\rangle }$ in Lemma 2.1. We first introduce five auxiliary registers

$$\begin{aligned} {|{0}\rangle }_{q_1} {|{0}\rangle }_{q_2} {|{0}\rangle }_{q_3} {|{0}\rangle }_a {|{0}\rangle }_b {|{0}\rangle }_c, \end{aligned}$$

(A1)

where register $q_i$ consists of 2n qubits storing the query result, register a is a qutrit with state space ${\mathbb {C}}^3=\text {span} \{{|{0}\rangle },{|{1}\rangle },{|{2}\rangle } \}$ used for generating ${|{\varphi (u)}\rangle }$ when u is an internal node, register b is a qudit with state space ${\mathbb {C}}^5=\text {span} \{{|{0}\rangle },\cdots ,{|{4}\rangle } \}$ storing conditions, and register c is a qubit used for generating ${|{\varphi (u)}\rangle }$ when u is one of the roots. Now $U_\varphi$ can be implemented as follows, where oracle query happens in the first and last step.

• 1. Query the oracle O on registers $r_1,q_1,q_2,q_3$ to obtain

  $$\begin{aligned} {|{u}\rangle }_{r_1} {|{\perp }\rangle }_{r_2} \bigotimes _{i=1}^{3} {|{\Gamma (u,i)}\rangle }_{q_i}\ {|{0}\rangle }_a {|{0}\rangle }_b {|{0}\rangle }_c. \end{aligned}$$

  (A2)

• 2. Apply the transformation ${|{q}\rangle }{|{b}\rangle } \mapsto {|{q}\rangle } {|{b+f(q)}\rangle }$ on register $q:=(q_1,q_2,q_3)$ and b. The function $f: \{0,1\}^{6n} \rightarrow \{0,\cdots ,4\}$ is defined as: $f(q)=0$ iff there’s no $q_i=\perp$, so that u is an internal node; $f(q)=i$ for $i=1,2,3$ iff there’s one and only one $q_i=\perp$, so that $u \in \{s,t\}$, and the ith register $q_i$ stores the value $\perp$; $f(q)=4$ iff there’s more than one $q_i=\perp$, so that $u\notin V(G_n)$. It can be easily seen that calculating f takes O(n) basic operations.

• 3. Conditioned on $b=0$, i.e. u is an internal node, apply the following two steps.

  • 3.1. Flip all the qubits of register $r_2$ so that it’s set to ${|{0^{2n}}\rangle }$. Apply quantum Fourier transform $QFT_3$ to register a, and then controlled by ${|{i}\rangle }_a,\, i\in \{0,1,2\}$, add (i.e. bit-wise modulo 2 addition) the value of register $q_{(i+1)}$ to register $r_2$, obtaining

    $$\begin{aligned} {|{u}\rangle }_{r_1} (\frac{1}{\sqrt{3}}\sum _{i=0}^2 {|{\Gamma (u,i+1)}\rangle }_{r_2} {|{i}\rangle }_a)\ \bigotimes _{i=0}^{2} {|{\Gamma (u,i+1)}\rangle }_{q_{(i+1)}} {|{b}\rangle }_b {|{0}\rangle }_c. \end{aligned}$$

    (A3)

    This controlled addition can be done in O(n) basic operations.

  • 3.2. Compare ${|{\Gamma (u,i)}\rangle }_{r_2}$ with ${|{\Gamma (u,j)}\rangle }_{q_j}$ for $j=1,2,3$ and subtract ${|{i}\rangle }_a$ with $(j-1)$, where j is the unique index j such that ${|{\Gamma (u,i)}\rangle }_{r_2}={|{\Gamma (u,j)}\rangle }_{q_j}$, obtaining

    $$\begin{aligned} {|{u}\rangle }_{r_1} {|{\varphi (u)}\rangle }_{r_2} \bigotimes _{i=1}^{3} {|{\Gamma (u,i)}\rangle }_{q_{i}}\ {|{0}\rangle }_a {|{b}\rangle }_b {|{0}\rangle }_c. \end{aligned}$$

    (A4)

    The uniqueness of the index j can be easily seen from the condition that all of u’s neighbours $\Gamma (u,i)$ are distinct. This compare (between binary strings) and subtract operation can be done in O(n) basic operations.

• 4. Conditioned on $b\in \{1,2,3\}$, i.e. u is one of the two roots and register $q_b$ stores $\perp$, apply the following steps.

  • 4.1. Swap register $q_b$ and $q_3$ so that the first two auxiliary registers store the genuine adjacent vertex name of $u \in \{s,t\}$. The conditioned SWAP operation can be done in O(n) basic operations.

  • 4.2. similar to step 3.1 and 3.2, transform ${|{\perp }\rangle }_{r_2}$ to ${|{\varphi (u)}\rangle }_{r_2}$ with the help of $H {|{0}\rangle }_c = \frac{1}{\sqrt{2}}({|{0}\rangle }+{|{1}\rangle })$.

  • 4.3. Repeat step 4.1 so that the order of register $q_i$ is restored, ensuring the success of step 6 and step 7.

• 5. Conditioned on $b = 4$, apply the identity transformation I, since register $r_1$ already stores ${|{\varphi (u)}\rangle }={|{\perp }\rangle }$ (the three ‘neighbours’ of $u\notin V(G_n)$ are all $\perp$).

• 6. Similar to step 2, apply the transformation ${|{q}\rangle }{|{b}\rangle } \mapsto {|{q}\rangle }{|{b-f(q)}\rangle }$ to register q, b, where the subtraction is modulo 5. Therefore, register b is recovered to ${|{0}\rangle }_b$.

• 7. Query the oracle O once more as in step 2, so that all the auxiliary registers are restored to zero.

Thus $U_\varphi$ can be implemented with 2 oracle queries and O(n) basic operations.

Spectral Decomposition of the Reduced Walk Matrix

In this Appendix, we prove Lemma 4.1 about the spectral decomposition of the reduced coined quantum walk matrix $M_U = M_S M_C$. The proof is inspired by [27], but we improved it with an easy observation of the relationship between monic and regular Chebyshev polynomial of the second kind (see Eq. (B47)), which allows us to obtain the analytical expression of $U_k$ shown in Eq. (B48). We will show in detail the complete proof for the sake of completeness and convenience of the readers.

We also expand or improve some of the implicit or complicated steps in [27], and point out a connection with another commonly used technique for analyzing quantum walk operators, i.e. the singular value decomposition, or more precisely, Jordan’s Lemma [37] about common invariant subspaces of two reflection operator, which has been used in Refs. [8, 10, 29, 38].

We first present the following helper Lemma B.1, which is implicit in [27] and similar to [32, Theorem 1], saying that in order to obtain the spectral decomposition of $M_U = M_S M_C$, we can instead consider the spectral decomposition of the following matrix

$$\begin{aligned} A^\dagger M_S A =: J_{2n}. \end{aligned}$$

(B5)

Lemma B.1

Consider the quantum walk operator $U=\textrm{Ref}_B\, \textrm{Ref}_A$, where $\textrm{Ref}_A = (2A A^\dagger -I)$ and A is a matrix with full column rank satisfying $A^\dagger A =I$. Let

$$\begin{aligned} {|{a}\rangle }:=A{|{v}\rangle },\ {|{b}\rangle }:= \textrm{Ref}_B {|{a}\rangle }, \end{aligned}$$

(B6)

where $\Vert {|{v}\rangle }\Vert ^2=1$. If

$$\begin{aligned} A^\dagger \textrm{Ref}_B A {|{v}\rangle } = \lambda {|{v}\rangle }. \end{aligned}$$

(B7)

Then, when $\left| \lambda \right| < 1$, we have

$$\begin{aligned} U {|{u_{+}}\rangle } = e^{i\varphi } {|{u_{+}}\rangle }, \quad U {|{u_{-}}\rangle } = e^{-i\varphi } {|{u_{-}}\rangle }, \end{aligned}$$

(B8)

where ${|{u_{\pm }}\rangle }:= {|{a}\rangle } -e^{\pm i\varphi } {|{b}\rangle }$, $\varphi := \arccos \lambda$, and $\Vert {|{u_{\pm }}\rangle }\Vert ^2 =2(1-\lambda ^2)$. And when $\lambda = \pm 1$, we have

$$\begin{aligned} U {|{a}\rangle } = \pm {|{a}\rangle }. \end{aligned}$$

(B9)

Proof

We first consider the case when $\left| \lambda \right| < 1$. From Eq. (B6) we know

$$\begin{aligned} U{|{a}\rangle }&= \textrm{Ref}_B (2AA^\dagger -I) A {|{v}\rangle } \end{aligned}$$

(B10)

$$\begin{aligned}&= \textrm{Ref}_B A {|{v}\rangle } \end{aligned}$$

(B11)

$$\begin{aligned}&= {|{b}\rangle }. \end{aligned}$$

(B12)

From Eq. (B7) we know

$$\begin{aligned} U {|{b}\rangle }&= \textrm{Ref}_B (2AA^\dagger -I) (\textrm{Ref}_B A {|{v}\rangle }) \end{aligned}$$

(B13)

$$\begin{aligned}&= 2\textrm{Ref}_B A \lambda {|{v}\rangle } -A{|{v}\rangle } \end{aligned}$$

(B14)

$$\begin{aligned}&= 2\lambda {|{b}\rangle } -{|{a}\rangle }. \end{aligned}$$

(B15)

Therefore, $\textrm{span} \{{|{a}\rangle },{|{b}\rangle }\}$ is an invariant subspace of U, and U takes the following matrix form:

$$\begin{aligned} L=\begin{bmatrix} 0 & -1 \\ 1 & 2\lambda \end{bmatrix}. \end{aligned}$$

(B16)

Let $\lambda =\cos \varphi$, then we obtain the eigenvalues and eigenvectors of L: $e^{\pm i\varphi }$ and $[1,-e^{\pm i\varphi }]^T$. This can be verified by the following identities:

$$\begin{aligned} 1-2\lambda e^{\pm i\varphi }&= 1 -2\cos (\pm \varphi ) e^{\pm i\varphi } \end{aligned}$$

(B17)

$$\begin{aligned}&= 1 -2\cos ^2(\pm \varphi ) -2i\cos (\pm \varphi ) \sin (\pm \varphi ) \end{aligned}$$

(B18)

$$\begin{aligned}&= -\cos (\pm 2\varphi ) -i\sin (\pm 2\varphi ) \end{aligned}$$

(B19)

$$\begin{aligned}&= - e^{\pm 2i\varphi } = e^{\pm i\varphi }\cdot (-e^{\pm i\varphi }). \end{aligned}$$

(B20)

Therefore, we obtain two eigenvalues $e^{\pm i\varphi }$ of U and their respective eigenvectors ${|{u_{\pm }}\rangle }:={|{a}\rangle } -e^{\pm i\varphi } {|{b}\rangle }$. We now calculate the square of its norm:

$$\begin{aligned} {\langle {u_{\pm }|u_{\pm }}\rangle }&=2-2\, \text {Re} ( e^{\pm i\varphi } {\langle {a|b}\rangle } ) \end{aligned}$$

(B21)

$$\begin{aligned}&= 2(1-\lambda ^2). \end{aligned}$$

(B22)

The second line follows from ${\langle {a|b}\rangle }={\langle {v}|} A^\dagger \text {Ref}_B A {|{v}\rangle } =\lambda$ and the fact that $A^\dagger \text {Ref}_B A$ is Hermitian whose eigenvalue $\lambda$ is a real number.

We now consider the case when $\lambda = \pm 1$. From Eq. (B7), we know ${\langle {v}|} A^\dagger \textrm{Ref}_B A {|{v}\rangle } = \pm 1$, and thus $\textrm{Ref}_B A {|{v}\rangle } = \pm A {|{v}\rangle }$. Therefore,

$$\begin{aligned} U {|{a}\rangle }&= U A {|{v}\rangle } \end{aligned}$$

(B23)

$$\begin{aligned}&= \textrm{Ref}_B (2AA^\dagger - I) A {|{v}\rangle } \end{aligned}$$

(B24)

$$\begin{aligned}&= \textrm{Ref}_B A {|{v}\rangle } \end{aligned}$$

(B25)

$$\begin{aligned}&= \pm A {|{v}\rangle } \end{aligned}$$

(B26)

$$\begin{aligned}&=\pm {|{a}\rangle }. \end{aligned}$$

(B27)

$\square$

Remark B.1

Suppose $\textrm{Ref}_B = 2BB^\dagger -I$. Let $D:=A^\dagger B$, and consider its singular value decomposition $\sum _i s_i {|{v_i}\rangle } {\langle {w_i}|}$, which is a common approach in Refs. [8, 10, 29, 38]. Then the connection between the eigenvalue $\lambda$ of $A^\dagger \, \textrm{Ref}_B\, A$ and the singular value s of D is:

$$\begin{aligned} \arccos (\lambda ) =2 \arccos (s). \end{aligned}$$

(B28)

Proof

Since $DD^\dagger {|{v_i}\rangle } =s_i^2 {|{v_i}\rangle }$, we have

$$\begin{aligned} A^\dagger \, \text {Ref}_B\, A&= A^\dagger (2BB^\dagger -I) A \end{aligned}$$

(B29)

$$\begin{aligned}&= 2(A^\dagger B) (A^\dagger B)^\dagger -I \end{aligned}$$

(B30)

$$\begin{aligned}&= 2DD^\dagger - I. \end{aligned}$$

(B31)

Therefore,

$$\begin{aligned} A^\dagger \, \text {Ref}_B\, A {|{v}\rangle } =\lambda {|{v}\rangle } \Leftrightarrow DD^\dagger {|{v}\rangle } = \frac{\lambda +1}{2} {|{v}\rangle }. \end{aligned}$$

(B32)

From the identity $\frac{\cos \phi _i +1}{2} = \cos ^2\frac{\phi _i}{2}$, we obtain Eq.(B28). $\square$

We now analyze the eigenvalues and eigenvectors of $J_{2n}=A^\dagger M_S A$. First, we will need the matrix expression of $J_{2n}$. Recall from Eq. (36) that the $2(2n+1)\times 2(n+1)$ matrix A is as follows:

$$\begin{aligned} A= \left[ \begin{array}{cccc|cccc} 1 & & & & & & & \\ & \sqrt{p} & & & & & & \\ & \sqrt{q} & & & & & & \\ & & \ddots & & & & & \\ & & & \sqrt{p} & & & & \\ & & & \sqrt{q} & & & & \\ \hline & & & & \sqrt{q} & & & \\ & & & & \sqrt{p} & & & \\ & & & & & \ddots & & \\ & & & & & & \sqrt{q} & \\ & & & & & & \sqrt{p} & \\ & & & & & & & 1 \end{array}\right] . \end{aligned}$$

(B33)

Since $M_S$ swaps two adjacent rows (columns), the ${2(n+1)\times 2(2n+1)}$ matrix $A^\dagger M_S$ is as follows:

$$\begin{aligned} A^\dagger M_S= \left[ \begin{array}{ccccccc|ccccccc} 0 & 1 & & & & & & & & & & & & \\ \sqrt{p} & 0 & 0 & \sqrt{q} & & & & & & & & & & \\ & & \sqrt{p} & 0 & 0 & \sqrt{q} & & & & & & & & \\ & & & & \ddots & & & & & & & & & \\ & & & & \sqrt{p} & 0 & 0 & \sqrt{q} & & & & & & \\ \hline & & & & & & \sqrt{q} & 0 & 0 & \sqrt{p} & & & & \\ & & & & & & & & & \ddots & & & & \\ & & & & & & & & \sqrt{q} & 0 & 0 & \sqrt{p} & & \\ & & & & & & & & & & \sqrt{q} & 0 & 0 & \sqrt{p} \\ & & & & & & & & & & & & 1 & 0 \\ \end{array}\right] . \end{aligned}$$

(B34)

Thus $J_{2n}$ is a ${2(n+1)\times 2(n+1)}$ tridiagonal matrix as shown below:

$$\begin{aligned} J_{2n}= \left[ \begin{array}{ccccc|ccccc} 0 & \sqrt{p} & & & & & & & & \\ \sqrt{p} & 0 & \sqrt{pq} & & & & & & \\ & \sqrt{pq} & 0 & \ddots & & & & & & \\ & & \ddots & \ddots & \sqrt{pq} & & & & & \\ & & & \sqrt{pq} & 0 & q & & & & \\ \hline & & & & q & 0 & \sqrt{pq} & & & \\ & & & & & \sqrt{pq} & \ddots & \ddots & & \\ & & & & & & \ddots & 0 & \sqrt{pq} & \\ & & & & & & & \sqrt{pq} & 0 & \sqrt{p} \\ & & & & & & & & \sqrt{p} & 0 \end{array} \right] . \end{aligned}$$

(B35)

I. Eigenvalues of $J_{2n}$

Denote by $|M|:= \det (M)$ the determinant of matrix M. We will calculate the characteristic equation $p(\lambda ):=|\lambda I -J_{2n}|$ of $J_{2n}$ as shown below:

$$\begin{aligned} p(\lambda ) =\left| \begin{array}{ccccc|ccccc} \lambda & -\sqrt{p} & & & & & & & & \\ -\sqrt{p} & \lambda & -\sqrt{pq} & & & & & & \\ & -\sqrt{pq} & \lambda & \ddots & & & & & & \\ & & \ddots & \ddots & -\sqrt{pq} & & & & & \\ & & & -\sqrt{pq} & \lambda & -q & & & & \\ \hline & & & & -q & \lambda & -\sqrt{pq} & & & \\ & & & & & -\sqrt{pq} & \ddots & \ddots & & \\ & & & & & & \ddots & \lambda & -\sqrt{pq} & \\ & & & & & & & -\sqrt{pq} & \lambda & -\sqrt{p} \\ & & & & & & & & -\sqrt{p} & \lambda \end{array} \right| . \end{aligned}$$

(B36)

We first introduce the following two principal submatrices. Here, we use the notation $J_{2n}[l:k]$ to represent the main sub-matrix of $J_{2n}$ formed by the rows and the columns with indices from l to k.

$$\begin{aligned} E_k&:= \lambda I - J_{2n}[1:k], \end{aligned}$$

(B37)

$$\begin{aligned} F_k&:= \lambda I -J_{2n}[2:k+1]. \end{aligned}$$

(B38)

Then we have the following recursive relations:

$$\begin{aligned} |E_2|&= \lambda |E_1| - p |E_0|, \end{aligned}$$

(B39)

$$\begin{aligned} |E_k|&=\lambda |E_{k-1}| -pq |E_{k-2}|,\ (3\le k \le n+1) \end{aligned}$$

(B40)

$$\begin{aligned} |E_k|&=\lambda |F_{k-1}| -p|F_{k-2}|,\ (2\le k \le n+1) \end{aligned}$$

(B41)

$$\begin{aligned} |F_k|&= \lambda |F_{k-1}| -pq |F_{k-2}|,\ (2\le k \le n) \end{aligned}$$

(B42)

where the first two terms are $|F_1|=\lambda ,\ |F_0|:=1$ and $|E_1|=\lambda ,\ |E_0|:=1$. Here, Eq. (B40) is obtained by expanding $|E_k|$ from its lower right corner, while Eq. (B41) is obtained by expanding $|E_k|$ from its upper left corner. Equation (B42) can be obtained by expanding $|F_k|$ from either from its upper left corner or its lower right corner. Dividing all the elements in $F_{k}$ by $\sqrt{pq}$ and denoting

$$\begin{aligned} |F_k|/\sqrt{pq}^k:= U_k(\lambda /\sqrt{pq}), \end{aligned}$$

(B43)

Eq. (B42) is now transformed to $U_k(\lambda /\sqrt{pq}) =\lambda /\sqrt{pq}\, U_{k-1}(\lambda /\sqrt{pq}) -U_{k-2}(\lambda /\sqrt{pq})$. If we let $x:=\frac{\lambda }{\sqrt{pq}}$, then Eq. (B42) further simplifies to

$$\begin{aligned} U_0(x)&=1, \ U_1(x)=x, \end{aligned}$$

(B44)

$$\begin{aligned} U_{k}(x)&= x\,U_{k-1}(x)-U_{k-2}(x). \end{aligned}$$

(B45)

Comparing the above equations with the recurrence relation of Chebyshev polynomial of the second kind:

$$\begin{aligned} {\tilde{U}}_0(x)&=1, \ {\tilde{U}}_1(x)=2x, \nonumber \\ {\tilde{U}}_k(x)&=2x {\tilde{U}}_{k-1}(x) -{\tilde{U}}_{k-2}(x), \ \text {for}\ k\ge 2, \end{aligned}$$

(B46)

we know:

$$\begin{aligned} U_k(x) = {\tilde{U}}_k(x/2). \end{aligned}$$

(B47)

From the general formula ${\tilde{U}}_k(\cos \theta ) = \frac{\sin (k+1)\theta }{\sin \theta }$, we let $\theta$ be such that it satisfies $\cos \theta := x/2 = \frac{\lambda }{2\sqrt{pq}}$. Therefore,

$$\begin{aligned} U_{k}(\lambda /\sqrt{pq}) = {\tilde{U}}_k(\cos \theta )=\frac{\sin (k+1)\theta }{\sin \theta }. \end{aligned}$$

(B48)

We now calculate $p(\lambda ) = |\lambda I -J_{2n}|$ (Eq. (B36)) by expanding its $(n+1)$-th row as follows. Denote by $E_{k}' = \lambda I -J_{2n}[2n+2-k+1:2n+2]$ the last k rows and columns of $\lambda I -J_{2n}$. It is easy to see that $|E_k'|=|E_k|$ from the centrosymmetry of $J_{2n}$.

$$\begin{aligned} p(\lambda )&= \sqrt{pq} \left| \begin{array}{ccc} E_{n-1} & & \\ -\sqrt{pq} & -\sqrt{pq} & \\ & -q & E_{n+1}' \end{array}\right| +\lambda \left| \begin{array}{cc} E_n & \\ & E_{n+1} \end{array}\right| +q \left| \begin{array}{ccc} E_{n} & -\sqrt{pq} & \\ & -q & -\sqrt{pq} \\ & & E_{n}' \end{array}\right| \end{aligned}$$

(B49)

$$\begin{aligned}&= -pq |E_{n-1}|\cdot |E_{n+1}| +\lambda |E_{n}|\cdot |E_{n+1}| -q^2 |E_{n}|^2 \end{aligned}$$

(B50)

$$\begin{aligned}&= (\lambda ^2 -q^2) |E_n|^2 -2pq\lambda |E_{n}|\cdot |E_{n-1}| +(pq)^2 |E_{n-1}|^2 \end{aligned}$$

(B51)

$$\begin{aligned}&= \big ((\lambda - q) |E_n| -pq |E_{n-1}|\big ) \big ((\lambda + q) |E_n| -pq |E_{n-1}|\big ). \end{aligned}$$

(B52)

Note that when choosing $-q$ as the pivot to expand the first determinant in Eq. (B49), the following sub-determinant is zero.

$$\begin{aligned} \left| \begin{array}{c|cc} E_{n-1} & & \\ -\sqrt{pq} & & \\ \hline & -\sqrt{pq} & E_n' \end{array}\right| = 0. \end{aligned}$$

(B53)

This is because the first n rows are rank deficient as there are only $n-1$ columns with non zero elements. The same reasoning applies to expanding the third determinant in Eq. (B49). The third line (Eq. (B51)) uses Eq. (B40) to further expand $|E_{n+1}|$.

We can further simplify the two components of $p(\lambda )$ in Eq. (B52) by using Eqs. (B40), (B41), (B42) to expand until $|F_{n-1}|,\ |F_{n-2}|$:

$$\begin{aligned}&(\lambda \mp q) |E_n| -pq |E_{n-1}| \end{aligned}$$

(B54)

$$\begin{aligned}&= (\lambda |E_n| -pq |E_{n-1}|) \mp q |E_n| \end{aligned}$$

(B55)

$$\begin{aligned}&= |E_{n+1}| \mp q |E_n| \end{aligned}$$

(B56)

$$\begin{aligned}&= \lambda |F_{n}| -p|F_{n-1}| \mp q(\lambda |F_{n-1}| -p|F_{n-2}|) \end{aligned}$$

(B57)

$$\begin{aligned}&= \lambda (\lambda |F_{n-1}| -pq |F_{n-2}|) +(\mp (1-p)\lambda -p) |F_{n-1}| \pm pq|F_{n-2}| \end{aligned}$$

(B58)

$$\begin{aligned}&= [\lambda (\lambda \mp 1) \pm p(\lambda \mp 1)]\cdot |F_{n-1}| -pq(\lambda \mp 1) |F_{n-2}| \end{aligned}$$

(B59)

$$\begin{aligned}&= (\lambda \mp 1)\ (|F_{n}| \pm p |F_{n-1}|). \end{aligned}$$

(B60)

Using Eq. (B43), we now obtain the eigenvalues of $J_{2n}$: $\pm 1$ and $\lambda _{\pm k}:= 2\sqrt{pq} \cos \theta _{\pm k}$, where $\lambda _{\pm k}$ are the 2n roots of the following equation:

$$\begin{aligned} \sqrt{q}\, U_n(\lambda /\sqrt{pq}) \pm \sqrt{p}\, U_{n-1}(\lambda /\sqrt{pq}) =0. \end{aligned}$$

(B61)

Combined with $U_{k}(\lambda /\sqrt{pq}) = \frac{\sin (k+1)\theta }{\sin \theta }$ shown by Eq. (B48), we obtain Eq. (41) in Lemma 4.1. It can also be seen that when $\theta _{k}$ corresponds to a root with ‘$+$’, then $\theta _{-k}:= \pi -\theta _k$ corresponds to a root with ‘−’, thus $\lambda _{-k} = -\lambda _k$.

II. Eigenvectors of $J_{2n}$

We now consider the eigenvectors of $J_{2n}$. Using Eqs. (B39), (B40), and “$(\lambda _{\pm k} \mp q) |E_n| -pq |E_{n-1}| = 0$” by Eq. (B54), it is not difficult to verify that the respective (unnormalized) eigenvector ${|{v_{\pm k}}\rangle }$ (which satisfies $(\lambda I -J_{2n}) {|{v_{\pm k}}\rangle }=0$, cf. Equation (B36) for $\lambda I -J_{2n}$) is

$$\begin{aligned} {[}1, \frac{|E_1|}{\sqrt{p}}, \frac{|E_2|}{\sqrt{p}\sqrt{pq}},\cdots , \frac{|E_{n-1}|}{\sqrt{p}\sqrt{pq}^{n-2}}, \frac{|E_n|}{\sqrt{p}\sqrt{pq}^{n-1}}, \pm (*)], \end{aligned}$$

(B62)

Where $(*)$ can be deduced from centrosymmetry. Combined with the relation of the sub-determinant $|E_k|$, $|F_k|$ shown in Eqs. (B41), (B42), the components and the square of the norm of the eigenvector ${|{v_{\pm k}}\rangle }$ can be calculated as follows.

1. $\lambda =\pm 1$

Since $\lambda ^2 =1$, we have

$$\begin{aligned} |F_k|&=\pm |F_{k-1}| -pq |F_{k-2}| \end{aligned}$$

(B63)

$$\begin{aligned}&= \pm (q+p)|F_{k-1}| -pq |F_{k-2}| \end{aligned}$$

(B64)

$$\begin{aligned}&= \pm q |F_{k-1}| \pm p(\pm |F_{k-2}|-pq|F_{k-3}|) -pq |F_{k-2}| \end{aligned}$$

(B65)

$$\begin{aligned}&= \pm q (|F_{k-1}| -p^2|F_{k-3}|) +p(1-q) |F_{k-2}| \end{aligned}$$

(B66)

$$\begin{aligned}&= \pm q (|F_{k-1}| -p^2|F_{k-3}|) +p^2 |F_{k-2}|. \end{aligned}$$

(B67)

Thus we know the difference between every other term, i.e. $|F_k|-p^2 |F_{k-2}|$, is a power series. Therefore,

$$\begin{aligned} |E_k|&= \pm |F_{k-1}| -p |F_{k-2}| \end{aligned}$$

(B68)

$$\begin{aligned}&= |F_{k}| -p^2 |F_{k-2}| \end{aligned}$$

(B69)

$$\begin{aligned}&= (\pm q)^{k-2} (|F_2| -p^2 |F_0|) \end{aligned}$$

(B70)

$$\begin{aligned}&= (\pm q)^{k-2}(1-pq -p^2) \end{aligned}$$

(B71)

$$\begin{aligned}&= (\pm 1)^k q^{k-1}. \end{aligned}$$

(B72)

The second line follows from substituting $q=1-p$ into Eq. (B42) and rearranging the terms. Thus the $(i+1)$-th component of ${|{v_{\pm 1}}\rangle }$ is

$$\begin{aligned} {\langle {i|v_{\pm 1}}\rangle }&=\frac{|E_i|}{\sqrt{p}\sqrt{pq}^{i-1}} \end{aligned}$$

(B73)

$$\begin{aligned}&= \frac{(\pm 1)^i q^{i-1}}{\sqrt{p}\sqrt{pq}^{i-1}} \end{aligned}$$

(B74)

$$\begin{aligned}&= (\pm 1)^i (\sqrt{{q}/{p}})^i /\sqrt{q}, \end{aligned}$$

(B75)

which corresponds to Eqs. (38), (39) in Lemma 4.1. Hence, the square of the norm of ${|{v_{\pm 1}}\rangle }$ is:

$$\begin{aligned} \Vert {|{v_{\pm 1}}\rangle }\Vert ^2&=2(1 +\frac{1}{q} \sum _{i=1}^{n} (\frac{q}{p})^i ) \end{aligned}$$

(B76)

$$\begin{aligned}&=2(1 + \frac{1}{q} \cdot \frac{q}{p} \cdot \frac{1-(q/p)^n}{1-q/p}) \end{aligned}$$

(B77)

$$\begin{aligned}&=2(1 + \frac{(q/p)^n - 1}{q-p}) \end{aligned}$$

(B78)

$$\begin{aligned}&= \frac{2}{q-p}( (q/p)^n -2p ), \end{aligned}$$

(B79)

which is Eq. (40) in Lemma 4.1.

2. $\lambda _{\pm k}$ for $k\in \{2,\dots ,n+1\}$

The $(i+1)$-th component of ${|{v_{\pm k}}\rangle }$ can be calculated as follows, where we omit the subscript “$\pm k$” for simplicity.

$$\begin{aligned} {\langle {i|v_\lambda }\rangle } =&\frac{\lambda |F_{i-1}| -p |F_{i-2}|}{\sqrt{p} \sqrt{pq}^{i-1}} \end{aligned}$$

(B80)

$$\begin{aligned}&=\frac{\lambda \sqrt{pq}^{i-1} U_{i-1}(\lambda /\sqrt{pq}) -p \sqrt{pq}^{i-2} U_{i-2}(\lambda /\sqrt{pq})}{\sqrt{p} \sqrt{pq}^{i-1}} \end{aligned}$$

(B81)

$$\begin{aligned}&= \frac{\lambda }{\sqrt{p}} U_{i-1}(\lambda /\sqrt{pq}) -\frac{1}{\sqrt{q}} U_{i-2}(\lambda /\sqrt{pq}), \end{aligned}$$

(B82)

which corresponds to Eqs. (46), (47) in Lemma 4.1, and the next Eq. (48) follows from Eq. (B48).

We now consider $\Vert {|{v_\lambda }\rangle } \Vert ^2$. We first calculate the square of the $(i+1)$-th component. For simplicity we denote $U_{i}(\lambda /\sqrt{pq})$ by $U_i$ in the following.

$$\begin{aligned} |{\langle {i|v_\lambda }\rangle }|^2&= \frac{\lambda ^2}{p} U_{i-1}^2 -\frac{2\lambda }{\sqrt{pq}} U_{i-1} U_{i-2} +\frac{1}{q} U_{i-2}^2 \end{aligned}$$

(B83)

$$\begin{aligned}&= \frac{\lambda ^2}{p} U_{i-1}^2 +(U_i^2 -\frac{\lambda ^2}{pq}U_{i-1}^2 -U_{i-2}^2) +\frac{1}{q} U_{i-2}^2 \end{aligned}$$

(B84)

$$\begin{aligned}&= \frac{p}{q}U_{i-2}^2 -\frac{\lambda ^2}{q} U_{i-1}^2 +U_i^2. \end{aligned}$$

(B85)

The second line is obtained by squaring both sides of the relation $U_i = \frac{\lambda }{\sqrt{pq}} U_{i-1} -U_{i-2}$ which follows from Eq. (B45). In order to make the relation true for $i\ge 1$, we set $U_{-1}:=0$. Then, using the identity $p/q -\lambda ^2/q +1 =(1-\lambda ^2)/q$, we have

$$\begin{aligned} \Vert {|{v_\lambda }\rangle } \Vert ^2/2&= 1 +\sum _{i=1}^n |{\langle {i|v_\lambda }\rangle }|^2 \end{aligned}$$

(B86)

$$\begin{aligned}&= 1+ 0 +\frac{p}{q} -\frac{\lambda ^2}{q} + \frac{1-\lambda ^2}{q} \sum _{i=1}^{n-2}U_i^2 -\frac{\lambda ^2}{q}U_{n-1}^2 +U_{n-1}^2 + U_{n}^2 \end{aligned}$$

(B87)

$$\begin{aligned}&=\frac{1-\lambda ^2}{q} +\frac{1-\lambda ^2}{q} \sum _{i=1}^{n-2}U_i^2 +(-\frac{\lambda ^2}{q}+1+\frac{p}{q})U_{n-1}^2 \end{aligned}$$

(B88)

$$\begin{aligned}&= \frac{1-\lambda ^2}{q} \sum _{i=0}^{n-1} U_i^2. \end{aligned}$$

(B89)

The third line uses Eq. (B61) satisfied by the eigenvalue $\lambda$. From the trigonometric expression of $U_i$ (Eq. (B48)), we have

$$\begin{aligned} \sum _{i=0}^{n-1} U_i^2&= \frac{1}{\sin ^2\theta } \sum _{i=0}^{n-1} \sin ^2(i+1)\theta \end{aligned}$$

(B90)

$$\begin{aligned}&=\frac{1}{\sin ^2\theta } \sum _{i=1}^{n} \frac{1-\cos i2\theta }{2} \end{aligned}$$

(B91)

$$\begin{aligned}&= \frac{1}{2\sin ^2\theta } (n- \frac{\sin n\theta \cdot \cos (n+1)\theta }{\sin \theta }) \end{aligned}$$

(B92)

$$\begin{aligned}&= \frac{1}{2\sin ^2\theta } (n \pm \sqrt{\frac{q}{p}} \frac{\sin 2(n+1)\theta }{2\sin \theta }). \end{aligned}$$

(B93)

Hence combined with $\Vert {|{u}\rangle }\Vert ^2 =2(1-\lambda ^2)$ in Lemma B.1 and the identity $\theta _{-k} =\pi -\theta _k$, we obtain Eq. (49) in Lemma 4.1. Note that the last line above uses Eq. (41) satisfied by $\theta$. The third line uses the identity obtained from comparing the real part of the following identities:

$$\begin{aligned}&\sum _{k=1}^n \cos kx +i\sin kx \end{aligned}$$

(B94)

$$\begin{aligned}&= \sum _{k=1}^n e^{ikx} = \frac{e^{ix}(e^{inx}-1)}{e^{ix}-1} \end{aligned}$$

(B95)

$$\begin{aligned}&= \frac{e^{ix} e^{\frac{inx}{2}} 2i \sin \frac{nx}{2}}{e^{\frac{ix}{2}} 2i\sin \frac{x}{2}} = \frac{\sin \frac{nx}{2}}{\sin \frac{x}{2}} e^{i\frac{(n+1)x}{2}} \end{aligned}$$

(B96)

$$\begin{aligned}&= \frac{\sin \frac{nx}{2}}{\sin \frac{x}{2}} (\cos \frac{(n+1)x}{2} +i\sin \frac{(n+1)x}{2}). \end{aligned}$$

(B97)